Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int MOD = 998244353;
long long int base = 29;
int MX = 1e9;
long double pi = 3.1415926535;
long long int n, q, kk, m;
vector<int> adj[100100];
int tin[100100];
int tout[100100];
int up[30][100100];
vector<int> ppl[20][100100];
vector<int> city[100100];
int logn;
int timer;
void prnt(vector<int> a) {
  for (int i = 0; i < a.size(); i++) cout << a[i] << " ";
  cout << endl;
}
vector<int> merg(vector<int> a, vector<int> b) {
  vector<int> c;
  vector<int> res;
  int i = 0, j = 0;
  while (i != a.size() && j != b.size()) {
    if (a[i] < b[j]) {
      c.push_back(a[i]);
      i++;
      while (i != a.size() && a[i] == a[i - 1]) i++;
    } else if (a[i] > b[j]) {
      c.push_back(b[j]);
      j++;
      while (j != b.size() && b[j] == b[j - 1]) j++;
    } else {
      c.push_back(a[i]);
      i++;
      j++;
      while (i != a.size() && a[i] == a[i - 1]) i++;
      while (j != b.size() && b[j] == b[j - 1]) j++;
    }
  }
  while (i != a.size()) {
    c.push_back(a[i]);
    i++;
    while (i != a.size() && a[i] == a[i - 1]) i++;
  }
  while (j != b.size()) {
    c.push_back(b[j]);
    j++;
    while (j != b.size() && b[j] == b[j - 1]) j++;
  }
  for (int i = 0; i < min((int)c.size(), 10); i++) res.push_back(c[i]);
  return res;
}
void dfs(int x, int p) {
  tin[x] = ++timer;
  up[0][x] = p;
  ppl[0][x] = merg(ppl[0][x], city[x]);
  for (int i = 1; i <= logn; i++) {
    up[i][x] = up[i - 1][up[i - 1][x]];
    ppl[i][x] = merg(ppl[i - 1][x], ppl[i - 1][up[i - 1][x]]);
  }
  for (int i = 0; i < adj[x].size(); i++) {
    int nod = adj[x][i];
    if (nod != p) dfs(nod, x);
  }
  tout[x] = ++timer;
}
bool is_parent(int x, int y) {
  return (tin[x] <= tin[y] && tout[x] >= tout[y]);
}
vector<int> lca(int a, int b) {
  vector<int> res;
  for (int i = logn; i >= 0; i--) {
    if (!is_parent(up[i][a], b)) {
      res = merg(res, ppl[i][a]);
      a = up[i][a];
    }
  }
  res = merg(res, city[a]);
  return res;
}
int lcaa(int a, int b) {
  if (is_parent(a, b)) return a;
  if (is_parent(b, a)) return b;
  for (int i = logn; i >= 0; i--) {
    if (!is_parent(up[i][a], b)) {
      a = up[i][a];
    }
  }
  return up[0][a];
}
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  cin >> n >> m >> q;
  while ((1 << logn) <= n) logn++;
  logn++;
  int x, y;
  for (int i = 0; i < n - 1; i++) {
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  for (int i = 1; i <= m; i++) {
    cin >> x;
    vector<int> temp;
    temp.push_back(i);
    city[x] = merg(city[x], temp);
  }
  dfs(1, 1);
  int z;
  for (int i = 1; i <= q; i++) {
    cin >> x >> y >> z;
    int lc = lcaa(x, y);
    vector<int> ans1 = lca(x, lc);
    vector<int> ans2 = lca(y, lc);
    vector<int> ans = merg(ans1, ans2);
    ans = merg(ans, city[lc]);
    cout << min((int)ans.size(), z);
    for (int i = 0; i < min((int)ans.size(), z); i++) cout << " " << ans[i];
    cout << endl;
  }
  return 0;
}